is_assignable and std::unique_ptr

橙三吉。 提交于 2019-12-22 05:36:18

问题


Here is a test file from gcc, live demo

struct do_nothing
{
    template <class T>
    void operator()(T*) {}
};

int
main()
{
    int i = 0;
    std::unique_ptr<int, do_nothing> p1(&i);
    std::unique_ptr<int> p2;
    static_assert(!std::is_assignable<decltype(p2), decltype(p1)>::value, ""); // note ! here.    
}

std::is_assignable

If the expression std::declval<T>() = std::declval<U>() is well-formed in unevaluated context, provides the member constant value equal true. Otherwise, value is false. Access checks are performed as if from a context unrelated to either type.

std::declval:

template<class T>
typename std::add_rvalue_reference<T>::type declval() noexcept;

The return type is T&& unless T is (possibly cv-qualified) void, in which case the return type is T.

Let's look at MoveAssignOnly:

struct MoveAssignOnly {
  MoveAssignOnly &operator=(MoveAssignOnly &) = delete;
  MoveAssignOnly &operator=(MoveAssignOnly &&) = default;
};

int main()
{
    static_assert(
    not std::is_assignable<MoveAssignOnly, MoveAssignOnly>::value, "");
}

live demo:

error: static_assert failed due to requirement '!std::is_assignable<MoveAssignOnly, MoveAssignOnly>::value'

Yes, it fails to compile because it provides a move assignment

Let's return to the gcc's test file and std::unique_ptr. As we know, std::unique_ptr also has move assignments.

However, unlike struct MoveAssignOnly, static_assert(!std::is_assignable<decltype(p2), decltype(p1)>::value, "");(more clearly, static_assert(!std::is_assignable<std::unique_ptr<int>, std::unique_ptr<int, do_nothing>>::value, ""); compiles happily.

I have struggled with libcxx's implementation of unique_ptr for long time, but still cannot figure out: how can std::unique_ptr be not assignable(! is_assignable) when std::unique_ptr provides move assignments?


回答1:


p1 and p2 are of a different type. Unlike with shared_ptr, the deleter of a unique_ptr is part of the pointer's type. This means the move assignment operator does not allow you to assign (even move-assign) between two unique_ptrs if their deleter types differ.

unique_ptr also offers an assignment operator template which allows assigning from an rvalue of unique_ptr with a different deleter, but the deleters must be assignable (see reference). So you can make your static assert fire by making the deleters assignable:

struct do_nothing
{
    template <class T>
    void operator()(T*) {}

    template <class T>
    operator std::default_delete<T>() { return {}; }
};

int
main()
{
    int i = 0;
    std::unique_ptr<int, do_nothing> p1(&i);
    std::unique_ptr<int> p2;
    static_assert(!std::is_assignable<decltype(p2), decltype(p1)>::value, ""); // note ! here.    
}

[Live example]



来源:https://stackoverflow.com/questions/53882498/is-assignable-and-stdunique-ptr

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